/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q8P Expand the following functions i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Expand the following functions in Legendre series.

Hint: Solve the recursion relation (5.8e)for Pl(x)and show that ∫a1 Pl(x) dx=1/2l+1 [Pl-1 (a)-Pl+1 (a)].

Short Answer

Expert verified

The expanded form of the function in the Legendre series is∫a 1 Pl(x) dx=1/2l+1 [Pl-1 (a)-Pl+1 (a)].

Step by step solution

01

Concept of Legendre series:

All the Legendre polynomials are orthogonal functions in the range (-1,1) and thus they satisfy the relation-

∫-11 Pl(x)Pm(x) dx = 0

(∀I≠ m)

02

Calculation to show that from recursion relation of the function ∫a1 Pl(x) dx=1/2l+1 [Pl-1 (a)-Pl+1 (a)]:

From the recursion relation:

(2l+1) Pl(x) = P'l+1 (x)-P'l-1 (x)

Pl(x) = [Pl+1(x) - Pl-1(x) ]/2l+1

∫a 1 Pl(x) dx= ∫a 1 [P'l+1 (a)-P'l-1 (x)]/2l+1 dx.

∫a 1 Pl(x) dx=1/2l+1 {Pl+1 (1)-Pl-1 (1) -Pl+1 (a) + Pl-1 (a)}

∫a 1 Pl(x) dx=1/2l+1 {Pl-1 (a)-Pl+1 (a)}

Hence, the given relation is proved.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To study the approximations in the table, a computer plots on the same axes the given function together with its small approximation and its asymptotic approximation. Use an interval large enough to show the asymptotic approximation agrees with the function for large . If the small approximation is not clear, plot it alone with the function over a small interval N3(x).

Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

|x|

Solve the differential equations in Problems 5 to 10 by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of are equal or differ by an integer, and in the latter case the larger gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is X times the solution you have, plus another Frobenius series, find the second solution.

x(x+1)y''-(x-1)y'+y=0

To study the approximations in the table, computer plot on the same axes the given function together with its small x approximation and its asymptotic approximation. Use an interval large enough to show the asymptotic approximation agreeing with the function for large x . If the small x approximation is not clear, plot it alone with the function over a small intervalJ3(x)

Verify that the differential equation in Problem11.13is not Fuchsian. Solve it by separation of variables to find the obvious solutiony=const. and a second solution in the form of an integral. Show that the second solution is not expandable in a Frobenius series.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.